Question

$A(2,5),B(-1,2)$ and $C(5,8)$ are co-ordinates of the vertices of the triangle $ABC.$ Points $P$ and $Q$ lie on $AB$ and $AC$ respectively, such that $AP:PB=AQ:QC=1:2$.
(i) Calculate the co-ordinate of $P$ and $Q.$
(ii) Show that :$PQ=\frac{1}{3}BC$.

Answer 1

$\left(i\right)$Given that $AP:PB=1:2$

Using section formula,

Section formula

$x=\frac{m{x}_2+n{x}_1}{m+n},$$y=\frac{m{y}_2+n{y}_1}{m+n}$

we have

Co-ordinates of $P$ are $P(x,y)=(\frac{1\times -1+2\times 2}{1+2},\frac{1\times 2+2\times 5}{1+2})=(\frac{-1+4}{3},\frac{2+10}{3})=(\frac{3}{3},\frac{12}{3})=(1,4)$

Given that $AQ:QC=1:2$

Using section formula, we have

Co-ordinates of $Q$ are $Q(x,y)=(\frac{1\times 5+2\times 2}{1+2},\frac{1\times 8+2\times 5}{1+2})=(\frac{5+4}{3},\frac{8+10}{3})=(\frac{9}{3},\frac{18}{3})=(3,6)$

$\left(ii\right)$Using distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

We have $BC=\sqrt{{(5+1)}^2+{(8-2)}^2}=\sqrt{36+36}=6\sqrt{2}$

$PQ=\sqrt{{(3-1)}^2+{(6-4)}^2}=\sqrt{4+4}=2\sqrt{2}$

$\frac{PQ}{BC}=\frac{2\sqrt{2}}{3\sqrt{2}}=\frac{1}{3}$

Hence $PQ=\frac{1}{3}BC$